Explanation Formula used: The Fourier series of a function on the interval ( − p , p ) is given as follows: f ( x ) = a 0 2 + ∑ n = 1 ∞ a n cos n x π p + b n sin n π p x Here, a 0 , a n and b n are defined as follows: a 0 = 1 p ∫ − p p f ( x ) d x a n = 1 p ∫ − p p f ( x ) ⋅ cos ( n x π p ) d x b n = 1 p ∫ − p p f ( x ) ⋅ sin ( n π p x ) d x Calculation: Use the above definition to obtain the Fourier expansion of the function f ( x ) = 2 x 2 − 1 , − 1 < x < 1 as follows: f ( x ) = a 0 2 + ∑ n = 1 ∞ a n cos n π x + b n sin n π x … … ( 1 ) Substitute f ( x ) = 2 x 2 − 1 and p = 1 in expression a 0 = 1 p ∫ − p p f ( x ) d x to obtain the value of a 0 as follows: a 0 = ∫ − 1 1 f ( x ) d x = ∫ − 1 1 ( 2 x 2 − 1 ) d x = [ 2 3 x 3 − x ] − 1 1 Further, solve the above expression: a 0 = 4 3 − 2 = 4 − 6 3 = − 2 3 Substitute f ( x ) = 2 x 2 − 1 and p = 1 in expression a n = 1 p ∫ − p p f ( x ) ⋅ cos ( n x π p ) d x to obtain the value of a n as follows: a n = ∫ − 1 1 f ( x ) ⋅ cos n x π d x = ∫ − 1 1 ( 2 x 2 − 1 ) ⋅ cos n π x d x = 2 ∫ − 1 1 x 2 ⋅ cos n π x d x − ∫ − 1 1 cos n π x d x = 2 [ x 2 n π sin n π x | − 1 1 − 2 n π ∫ − 1 1 x sin π x d x ] − [ sin n π x n π ] − 1 1 Further solve the above expression: a n = 2 [ ( 0 ) − 2 n π ∫ − 1 1 x sin π x d x ] − ( 0 ) ( sin n π x = 0 , n ∈ Z

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ISBN: 9781284127003

Author: ZILL

Publisher: JONES+BARTLETT LEARNING,LLC (CC)

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The expansion of function f ( x ) = 2 x 2 − 1 , − 1 < x < 1 using Fourier series.

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The expansion of function f(x)=2x2−1, −1<x<1 using Fourier series.

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Chapter 12 Solutions

The expansion of function f ( x ) = 2 x 2 − 1 , − 1 &lt; x &lt; 1 using Fourier series.

Solution Summary: The author explains the Fourier expansion of a function f(x)=2x,-11 using fourier series.

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The expansion of function f(x)=2x2−1, −1<x<1 using Fourier series.

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Chapter 12 Solutions

The expansion of function f ( x ) = 2 x 2 − 1 , − 1 < x < 1 using Fourier series.